Observing this limit $$\lim_{n\to \infty} \left(n-{\ln(n!)\over \sqrt[n]{n^k\ln2 \ln3\ln4\cdots\ln n}}\right)=2k$$

I did try a few quite large values of n; the limit seem to converge, but slow.

Can this limit be correct and how can I shows it?

  • $\begingroup$ Note that $\ln(n!)$ is $(n-1)$ times arithmetic mean of $A=\{\ln 2,\ldots,\ln n\}$ and $\sqrt[n]{\ln2\ln3\cdots\ln n}$ is related to geometric mean of numbers in $A$ $\endgroup$ – Qurultay Mar 26 '18 at 18:20
  • $\begingroup$ Not sure how you came up with this guess, but my numerical computation tells that the limit diverges to $-\infty$ regardless of the value of $k$. $\endgroup$ – Sangchul Lee Mar 26 '18 at 18:56
  • $\begingroup$ Thank you @Sangchul Lee. $\endgroup$ – user545955 Mar 26 '18 at 19:12

Let me provide some computation that concludes the question. With a bit of effort, one can show that there exists a constant $C > 0$ satisfying

$$ n - \frac{\log(n!)}{(n^k \prod_{j=2}^{n} \log j)^{1/n}} \leq - \operatorname{li}(n) - n \log\left(1 - \frac{1}{\log n}\right) + k \log n + \log\log n + C $$

for all $n \geq 2$ and $k \in \mathbb{R}$, where $\operatorname{li}(x)$ is the logarithmic integral function. Then it is not hard to check that this bound behaves like $-n^{1-o(1)}$ and hence the limit diverges.


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